Explanation:
<h3>MgCl2, CaCl2, MgO, Na2S, CaH2, AlF3, NaH, KH, K2O, KI, RbCl, NaBr, CaH2 etc.</h3>
<h2>hope it helps.</h2><h2>stay safe healthy and happy..</h2>
Answer:
8.73
Explanation:
The concentration of acetic acid can be determined as follows:






Moles of
= 
=0.0090 moles
Moles of 
= 0.0090 moles
The equation for the reaction can be expressed as :
----->

Concentration of
ion = 
= 
= 0.052 M
Hydrolysis of
ion:
----->


⇒ 
= 
As K is so less, then x appears to be a very infinitesimal small number
0.052-x ≅ x





![[OH] = x =0.535*10^{-5}](https://tex.z-dn.net/?f=%5BOH%5D%20%3D%20x%20%3D0.535%2A10%5E%7B-5%7D)
![pOH = -log[OH^-]](https://tex.z-dn.net/?f=pOH%20%3D%20-log%5BOH%5E-%5D)
![pOH = -log[0.535*10^{-5}]](https://tex.z-dn.net/?f=pOH%20%3D%20-log%5B0.535%2A10%5E%7B-5%7D%5D)

pH = 14 - pOH
pH = 14 - 5.27
pH = 8.73
Hence, the pH of the titration mixture = 8.73
Answer:
If you're looking at the data as a whole, it would most likely be 100ml.
Explanation: The definition of precise is data close together so 100ml is furthest away from the other recorded numbers
Answer : The mass of helium gas in the balloon is, 1.23 grams.
Explanation : Given,
Volume of helium gas = 6.9 L
First we have to calculate moles of helium gas at STP.
As we know that, 1 mole of substance occupy 22.4 L volume of gas.
As, 22.4 L volume of helium gas present in 1 mole of helium
So, 6.9 L volume of helium gas present in
of helium
Now we have to calculate the mass of helium gas.

Molar mass of He gas = 4 g/mol

Thus, the mass of helium gas in the balloon is, 1.23 grams.
Explanation:
Because molarity is classified as moles of solute per liter of water, dilution of the water may result in a reduction of its concentration.
Therefore, because the amount of moles of solute has to be constant for dilution, you will use the molarity and volume of that same target solution to calculate how many moles of solute will be present in the sample of the stock solution that you dilute.
c = 
⇒ n
=
c
⋅ V
= 0.250 M ⋅ 6.00 L =
1.5 moles HCl
Now all you have to do is figure out what volume of 6.0 M stock solution will contain 1.5 moles of hydrochloric acid
c = 
V = 
=
=
0.25 L
Expressed in milliliters, the answer will be
→ rounded to two sig figs